1. Introduction
It is well known that the nonlinear Schrödinger (NLS) equation [1]1.1 ) can describe the propagation of waves in Kerr-type media [2]. Although the NLS equation (1.1 ) can accurately describe the propagation of pulse waves in the picosecond range [3], experiments in the sub-picosecond (i.e. femtosecond) range and short pulses of high-intensity require consideration of higher-order correction terms [4]. Based on this physical reason, Kodama and Hasegawa [5] considered an NLS equation with high-order correction terms, which is also named the high-order (HNLS) NLS equation, it is as follows:1.2 ) also has the following four known types of nonlinear integrable models:
$\begin{eqnarray}{\rm{i}}{q}_{t}+\displaystyle \frac{1}{2}{q}_{{zz}}+| q{| }^{2}q=0,\end{eqnarray}$
is an important nonlinear integrable model with important applications in theoretical physics such as plasma, fluid mechanics, and nonlinear optics. For instance, in nonlinear optics, the NLS equation ( $\begin{eqnarray}\begin{array}{l}{\rm{i}}{q}_{t}+\alpha ({q}_{{zz}}+2| q{| }^{2}q)\\ \quad +\,{\rm{i}}\beta ({{aq}}_{{zzz}}+b| q{| }^{2}{q}_{z}+{cq}| q{| }_{z}^{2})=0,\end{array}\end{eqnarray}$
where q = q(z, t) is a complex valued function of the real variables z and t, α, β and a, b, c are real constants. In fact, in addition to the NLS equation at α = 1, β = 0, equation (If $\alpha =\tfrac{1}{2},\beta =1$ and a: b: c = 0: 1: 1, equation (
$\begin{eqnarray}{\rm{i}}{q}_{t}+\displaystyle \frac{1}{2}{q}_{{zz}}+| q{| }^{2}q+{\rm{i}}(| q{| }^{2}{q}_{z}+q| q{| }_{z}^{2})=0.\end{eqnarray}$
If $\alpha =\tfrac{1}{2},\beta =1$ and a: b: c = 0: 1: 0, equation (
$\begin{eqnarray}{\rm{i}}{q}_{t}+\displaystyle \frac{1}{2}{q}_{{zz}}+| q{| }^{2}q+{\rm{i}}| q{| }^{2}{q}_{z}=0.\end{eqnarray}$
If $\alpha =\tfrac{1}{2},\beta =1$ and a: b: c = 1: 6: 3, equation (
$\begin{eqnarray}\begin{array}{l}{\rm{i}}{q}_{t}+\displaystyle \frac{1}{2}{q}_{{zz}}+| q{| }^{2}q\\ \,+\,{\rm{i}}({q}_{{zzz}}+6| q{| }^{2}{q}_{z}+3q| q{| }_{z}^{2})=0.\end{array}\end{eqnarray}$
If $\alpha =\tfrac{1}{2},\beta =1$ and a: b: c = 1: 6: 0, equation (
$\begin{eqnarray}{\rm{i}}{q}_{t}+\displaystyle \frac{1}{2}{q}_{{zz}}+| q{| }^{2}q+{\rm{i}}({q}_{{zzz}}+6| q{| }^{2}{q}_{z})=0.\end{eqnarray}$
Recently, nonlocal integrable equations have become a research hotspot in soliton theory due to their ability to explain complex phenomena occurring in multi-position systems and explore new physical applications [10–24]. For example, in physical applications, the nonlocal modified Korteweg–de Vries (mKdV) equation related to the Alice-Bob system has parity in displacement and antisymmetry in delay time [25, 26]. By using a particular solution of the nonlocal mKdV equation, people can theoretically obtain the characteristics of two related dipole-blocking events in atmospheric and oceanic dynamic systems. In 2019, Cen et al [27] obtained the following reverse space-time nonlocal Hirota equation by discussing the zero curvature condition or AKNS equation for a new class of integrable systems1.7 ). Subsequently, Zuo and Zhang [28] used the bilinear derivative method to construct exact solutions for the nonlocal Hirota equation (1.7 ). In 2022, Xia et al [29] constructed soliton solutions for the nonlocal Hirota equation (1.7 ) using the N-fold Darboux transformation method and provided corresponding dynamic behavior analysis. Meanwhile, Zhuang et al constructed multi-soliton solutions for the nonlocal Hirota equation (1.7 ) using the Riemann–Hilbert (RH) method [30]. Specially, Peng and Chen [31] used the RH method and physics-informed neural networks (PINN) to discuss N-double pole solutions for the nonlocal Hirota equation (1.7 ) with nonzero boundary conditions etc [32–34].
$\begin{eqnarray}\begin{array}{l}{\rm{i}}{q}_{t}(z,t)+\alpha ({q}_{{zz}}(z,t)-2{q}^{2}(z,t)\bar{q}(-z,-t))\\ \quad +\,{\rm{i}}\beta ({q}_{{zzz}}(z,t)-6q(z,t)\bar{q}(-z,-t){q}_{z}(z,t))\\ \quad =\,0,\,\alpha ,\beta \in {\mathbb{R}},\end{array}\end{eqnarray}$
and they used the Darboux–Crum transformation method to obtain multiple types of solutions for the nonlocal Hirota equation (With the rapid development of soliton theory, the RH method has become a powerful tool for studying nonlinear partial differential equations. It can be well used to analyze local and nonlocal integrable equations, such as the initial value problems [35–38], the initial boundary value problems [39–44], the soliton solutions [45–47], and the large time asymptotic behavior [48–53]. In this paper, we will use the RH method to discuss the Cauchy problem of the nonlocal Hirota equation (1.7 ) with the following step-like initial value
$\begin{eqnarray}{q}_{0}(z)\to 0,\,z\to -\infty ,\end{eqnarray}$
$\begin{eqnarray}{q}_{0}(z)\to \delta ,\,z\to \infty ,\end{eqnarray}$
sufficiently fast, where q0(z) = q(z, 0) and δ > 0 is an arbitrary constant. In addition, we assume that the solution q(z, t) of Cauchy problem admits the following boundary conditions for all t > 0: $\begin{eqnarray}q(z,t)=o(1),\,z\to -\infty ,\end{eqnarray}$
$\begin{eqnarray}q(z,t)=\delta +o(1),\,z\to \infty .\end{eqnarray}$
Our paper is organized as follows. In section 2 , we present the Jost function solutions for the large z behavior of the Lax pair equations. In section 3 , we investigate the properties of the spectral functions v(λ) and uj(λ), j = 1, 2. In section 4 , we establish the RH problem to represent the solution of the Cauchy problem (1.7 ), (1.8a )–(1.9b ), and obtain the exact solution of the nonlocal Hirota equation (1.7 ) in a special case. In section 5 , we give our conclusions and discussions.
2. The spectral analysis
Similar to [29, 30], we consider the matrix Lax pair of the reverse space-time nonlocal Hirota equation (1.7 ) as follows
$\begin{eqnarray}{\varphi }_{z}=M(z,t,\lambda )\varphi =(-{\rm{i}}\lambda J+Q)\varphi ,\end{eqnarray}$
$\begin{eqnarray}{\varphi }_{t}=N(z,t,\lambda )\varphi =[-(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})J+P]\varphi ,\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}J & = & \left(\begin{array}{cc}1 & 0\\ 0 & -1\end{array}\right),Q(z,t)=\left(\begin{array}{cc}0 & q(z,t)\\ \bar{q}(-z,-t) & 0\end{array}\right),\\ P(z,t,\lambda ) & = & \left(\begin{array}{cc}A & B\\ C & -A\end{array}\right),\end{array}\end{eqnarray}$
with $\begin{eqnarray*}\begin{array}{rcl}A & = & -2{\rm{i}}\beta q(z,t)\bar{q}(-z,-t)\lambda -{\rm{i}}\alpha q(z,t)\bar{q}(-z,-t)\\ & & +\beta ({q}_{z}(z,t)\bar{q}(-z,-t)-q(z,t){\bar{q}}_{z}(-z,-t)),\\ B & = & 4\beta q(z,t){\lambda }^{2}+(2{\rm{i}}\beta {q}_{z}(z,t)+2\alpha q(z,t))\lambda \\ & & +{\rm{i}}\alpha {q}_{z}(z,t)-\beta ({q}_{{zz}}(z,t)-2{q}^{2}(z,t)\bar{z}(-z,-t)),\\ C & = & 4\beta \bar{q}(-z,-t){\lambda }^{2}+(-2{\rm{i}}\beta {\bar{q}}_{z}(-z,-t)\\ & & +2\alpha \bar{q}(-z,-t))\lambda \\ & & -{\rm{i}}\alpha {\bar{q}}_{z}(-z,-t)+\beta (-{\bar{q}}_{{zz}}(-z,-t)\\ & & +2q(z,t){\bar{q}}^{2}(-z,-t)),\end{array}\end{eqnarray*}$
where $\lambda \in {\mathbb{C}}$ is a spectral parameter, α and β are real constants. Then, we have $\begin{eqnarray}{\varphi }_{z}+{\rm{i}}\lambda J\varphi =Q\varphi ,\end{eqnarray}$
$\begin{eqnarray}{\varphi }_{t}+(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})J\varphi =P\varphi .\end{eqnarray}$
To obtain the asymptotic spectrum problem, on the one hand, we introduce the following symbols: $\begin{eqnarray}\begin{array}{rcl}{Q}_{+} & = & \left(\begin{array}{cc}0 & \delta \\ 0 & 0\end{array}\right),\,{Q}_{-}=\left(\begin{array}{cc}0 & 0\\ \delta & 0\end{array}\right),\\ {P}_{+} & = & \left(\begin{array}{cc}0 & (4\beta {\lambda }^{2}+2\alpha \lambda )\delta \\ 0 & 0\end{array}\right),\\ {P}_{-} & = & \left(\begin{array}{cc}0 & 0\\ (4\beta {\lambda }^{2}+2\alpha \lambda )\delta & 0\end{array}\right).\end{array}\end{eqnarray}$
It is easy to obtain that $\begin{eqnarray}Q(z,t)\to {Q}_{\pm },\,P(z,t,\lambda )\to {P}_{\pm },\,z\to \pm \infty .\end{eqnarray}$
On the other hand, we introduce the following symbols, similarly $\begin{eqnarray}{U}_{+}=-{\rm{i}}\lambda J+{Q}_{+}=\left(\begin{array}{cc}-{\rm{i}}\lambda & \delta \\ 0 & {\rm{i}}\lambda \end{array}\right),\end{eqnarray}$
$\begin{eqnarray}{U}_{-}=-{\rm{i}}\lambda J+{Q}_{-}=\left(\begin{array}{cc}-{\rm{i}}\lambda & 0\\ \delta & {\rm{i}}\lambda \end{array}\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{V}_{+} & = & -(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})J+{P}_{+}\\ & = & \left(\begin{array}{cc}-(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2}) & (4\beta {\lambda }^{2}+2\alpha \lambda )\delta \\ 0 & 4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2}\end{array}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{V}_{-} & = & -(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})J+{P}_{-}\\ & = & \left(\begin{array}{cc}-(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2}) & 0\\ (4\beta {\lambda }^{2}+2\alpha \lambda )\delta & 4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2}\end{array}\right),\end{array}\end{eqnarray}$
naturally, we have $\begin{eqnarray}{U}_{\pm }(z,t,\lambda )={H}_{\pm }(\lambda )(-{\rm{i}}\lambda J){H}_{\pm }^{-1}(\lambda ),\end{eqnarray}$
$\begin{eqnarray}{V}_{\pm }(z,t,\lambda )={H}_{\pm }(\lambda )[-(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})J]{H}_{\pm }^{-1}(\lambda ),\end{eqnarray}$
and $\begin{eqnarray}{\varphi }_{\pm }(z,t,\lambda )={H}_{\pm }(\lambda ){{\rm{e}}}^{-[{\rm{i}}\lambda z+(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})t]J},\end{eqnarray}$
where $\begin{eqnarray}{H}_{+}(\lambda )=\left(\begin{array}{cc}1 & \displaystyle \frac{\delta }{2{\rm{i}}\lambda }\\ 0 & 1\end{array}\right),\quad {H}_{-}(\lambda )=\left(\begin{array}{cc}1 & 0\\ \displaystyle \frac{\delta }{2{\rm{i}}\lambda } & 1\end{array}\right).\end{eqnarray}$
For the linear spectrum problem equations (2.3a )–(2.3b ), we make the function transformation as follows2.12a )–(2.12b ) can be written in fully differential form
$\begin{eqnarray}{\phi }_{\pm }(z,t,\lambda )={\varphi }_{\pm }(z,t,\lambda ){{\rm{e}}}^{[{\rm{i}}\lambda z+(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})t]J},\end{eqnarray}$
then, we have $\begin{eqnarray}{\phi }_{\pm }\to {H}_{\pm },\quad {as}\quad z\to \pm \infty ,\end{eqnarray}$
and obtain asymptotic Lax pair as follows: $\begin{eqnarray}{\left({H}_{\pm }^{-1}{\phi }_{\pm }\right)}_{z}+{\rm{i}}\lambda [J,{H}_{\pm }^{-1}{\phi }_{\pm }]={H}_{\pm }^{-1}{\rm{\Delta }}{Q}_{\pm }{\phi }_{\pm },\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\left({H}_{\pm }^{-1}{\phi }_{\pm }\right)}_{t}+(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})[J,{H}_{\pm }^{-1}{\phi }_{\pm }]\\ \quad =\,{H}_{\pm }^{-1}{\rm{\Delta }}{P}_{\pm }{\phi }_{\pm },\end{array}\end{eqnarray}$
where $\begin{eqnarray*}{\rm{\Delta }}{Q}_{\pm }=Q-{Q}_{\pm }\quad {\rm{and}}\quad {\rm{\Delta }}{P}_{\pm }=P-{P}_{\pm }.\end{eqnarray*}$
Therefore, the asymptotic spectrum problem equations ( $\begin{eqnarray}\begin{array}{l}{\rm{d}}({{\rm{e}}}^{{\rm{i}}\omega (\lambda )\hat{J}}{H}_{\pm }^{-1}{\phi }_{\pm })={{\rm{e}}}^{{\rm{i}}\omega (\lambda )\hat{J}}\\ \,\times \,[{H}_{\pm }^{-1}({\rm{\Delta }}{Q}_{\pm }{\rm{d}}z+{\rm{\Delta }}{P}_{\pm }{\rm{d}}t){\phi }_{\pm }],\end{array}\end{eqnarray}$
where $\hat{J}$ is a matrix operator (see [42]) and ω(λ) = λz + (4βλ3 + 2αλ2)t.Noting φ− = φ1 and φ+ = φ2, and integrating equation (2.13 ) along two special paths (– ∞ , t) → (z, t) and (+ ∞ , t) → (z, t), we can obtain two Jost function solutions for equations (2.12a )–(2.12b ) as follows
$\begin{eqnarray}\begin{array}{l}{\phi }_{1}(z,t,\lambda )={H}_{-}(\lambda )+{\displaystyle \int }_{-\infty }^{z}{D}_{-}(z,x,t,\lambda )\\ \quad \times \,{\rm{\Delta }}{Q}_{-}(x,t,\lambda ){\phi }_{1}(x,t,\lambda ){{\rm{e}}}^{{\rm{i}}\lambda (z-x)J}{\rm{d}}x,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\phi }_{2}(z,t,\lambda )={H}_{+}(\lambda )+{\displaystyle \int }_{+\infty }^{z}{D}_{+}(z,x,t,\lambda )\\ \quad \times {\rm{\Delta }}{Q}_{+}(x,t,\lambda ){\phi }_{2}(x,t,\lambda ){e}^{{\rm{i}}\lambda (z-x)J}{\rm{d}}x,\end{array}\end{eqnarray}$
where ${D}_{\pm }(z,x,t,\lambda )={\varphi }_{\pm }(z,t,\lambda ){\left[{\varphi }_{\pm }\left(x,t,\lambda \right)\right]}^{-1}$.Let ${\phi }_{j}(z,t,\lambda )=({\phi }_{j}^{\left(1\right)}(z,t,\lambda ),{\phi }_{j}^{\left(2\right)}(z,t,\lambda ))$, $j=1,2$, then the column vectors ${\phi }_{j}^{\left(1\right)}(z,t,\lambda )$ and ${\phi }_{j}^{\left(2\right)}(z,t,\lambda )$ meet the following properties:
(i) ${\phi }_{1}^{\left(1\right)}(z,t,\lambda )$ and ${\phi }_{2}^{\left(2\right)}(z,t,\lambda )$ are continuous in $\lambda \in {\overline{{\mathbb{C}}}}_{+}\setminus \{0\}$ and are analytical and well-defined for $\lambda \in {{\mathbb{C}}}_{+}$.
(ii) ${\phi }_{1}^{\left(2\right)}(z,t,\lambda )$ and ${\phi }_{2}^{\left(1\right)}(z,t,\lambda )$ are continuous in $\lambda \in {\overline{{\mathbb{C}}}}_{-}$ and are analytical and well-defined for $\lambda \in {{\mathbb{C}}}_{-}$.
(iii) The asymptotic properties for $\lambda \to \infty $
$\begin{eqnarray}{\phi }_{1}^{\left(1\right)}(z,t,\lambda )=\left(\begin{array}{c}1\\ 0\end{array}\right)+O\left(\displaystyle \frac{1}{\lambda }\right),\,\lambda \in {{\mathbb{C}}}_{+},\end{eqnarray}$
$\begin{eqnarray}{\phi }_{2}^{\left(2\right)}(z,t,\lambda )=\left(\begin{array}{c}0\\ 1\end{array}\right)+O\left(\displaystyle \frac{1}{\lambda }\right),\,\lambda \in {{\mathbb{C}}}_{+},\end{eqnarray}$
$\begin{eqnarray}{\phi }_{1}^{\left(2\right)}(z,t,\lambda )=\left(\begin{array}{c}0\\ 1\end{array}\right)+O\left(\displaystyle \frac{1}{\lambda }\right),\,\lambda \in {{\mathbb{C}}}_{-},\end{eqnarray}$
$\begin{eqnarray}{\phi }_{2}^{\left(1\right)}(z,t,\lambda )=\left(\begin{array}{c}1\\ 0\end{array}\right)+O\left(\displaystyle \frac{1}{\lambda }\right),\,\lambda \in {{\mathbb{C}}}_{-}.\end{eqnarray}$
(iv) The determinant properties
$\begin{eqnarray*}\begin{array}{l}\det {\phi }_{j}(z,t,\lambda )=1,\\ j=1,2,z\in {\mathbb{R}},t\geqslant 0,\lambda \in {\mathbb{R}}.\end{array}\end{eqnarray*}$
(v) The matrix-valued functions ${\varphi }_{j}(z,t,\lambda ),j\,=\,1,2$ represented by
$\begin{eqnarray}{\varphi }_{j}(z,t,\lambda )={\phi }_{j}(z,t,\lambda ){{\rm{e}}}^{-{\rm{i}}\omega (\lambda )J},\lambda \in {\mathbb{R}}\setminus \{0\},\end{eqnarray}$
admit the boundary conditions as follows: $\begin{eqnarray}{\varphi }_{1}(z,t,\lambda )\to {\varphi }_{-}(z,t,\lambda ),\lambda \to -\infty ,\end{eqnarray}$
$\begin{eqnarray}{\varphi }_{2}(z,t,\lambda )\to {\varphi }_{+}(z,t,\lambda ),\lambda \to +\infty ,\end{eqnarray}$
and they are also Jost function solutions of equations ((vi) The symmetry properties
$\begin{eqnarray}E\overline{{\phi }_{1}(-z,-t,-\overline{\lambda })}{E}^{-1}={\phi }_{2}(z,t,\lambda ),\lambda \in {\mathbb{R}}\setminus \{0\},\end{eqnarray}$
where $E=\left(\begin{array}{cc}0 & 1\\ 1 & 0\end{array}\right).$ (vii) As $\lambda \to 0$, ${\phi }_{j}^{\left(1\right)}(z,t,\lambda )$ and ${\phi }_{j}^{\left(2\right)}(z,t,\lambda )$ have the following asymptotic properties
$\begin{eqnarray}{\phi }_{1}^{\left(1\right)}(z,t,\lambda )=\displaystyle \frac{1}{\lambda }\left(\begin{array}{c}{m}_{1}(z,t)\\ {m}_{2}(z,t)\end{array}\right)+O(1),\end{eqnarray}$
$\begin{eqnarray}{\phi }_{1}^{\left(2\right)}(z,t,\lambda )=\displaystyle \frac{2{\rm{i}}}{\delta }\left(\begin{array}{c}{m}_{1}(z,t)\\ {m}_{2}(z,t)\end{array}\right)+O(\lambda ),\end{eqnarray}$
$\begin{eqnarray}{\phi }_{2}^{\left(1\right)}(z,t,\lambda )=-\displaystyle \frac{2{\rm{i}}}{\delta }\left(\begin{array}{c}\overline{{m}_{2}}(-z,-t)\\ \overline{{m}_{1}}(-z,-t)\end{array}\right)+O(\lambda ),\end{eqnarray}$
$\begin{eqnarray}{\phi }_{2}^{\left(2\right)}(z,t,\lambda )=-\displaystyle \frac{1}{\lambda }\left(\begin{array}{c}\overline{{m}_{2}}(-z,-t)\\ \overline{{m}_{1}}(-z,-t)\end{array}\right)+O(1),\end{eqnarray}$
where ${m}_{1}(z,t)$ and ${m}_{2}(z,t)$ satisfy the following linear Volterra integral equation $\begin{eqnarray}{m}_{1}(z,t)={\int }_{-\infty }^{z}q(x,t){m}_{2}(x,t){\rm{d}}x,\end{eqnarray}$
$\begin{eqnarray}{m}_{2}(z,t)=-\displaystyle \frac{{\rm{i}}\delta }{2}+{\int }_{-\infty }^{z}(\bar{q}(-x,-t)-2\delta ){m}_{1}(x,t){\rm{d}}x.\end{eqnarray}$
Noting the construction of ${\{{\phi }_{j}(z,t,\lambda )\}}_{1}^{2}$ given by equations (
$\begin{eqnarray}{\phi }_{1}^{\left(1\right)}(z,t,\lambda )=\displaystyle \frac{1}{\lambda }\left(\begin{array}{c}{m}_{1}(z,t)\\ {m}_{2}(z,t)\end{array}\right)+O(1),\end{eqnarray}$
$\begin{eqnarray}{\phi }_{1}^{\left(2\right)}(z,t,\lambda )=\left(\begin{array}{c}\widetilde{{m}_{1}}(z,t)\\ \widetilde{{m}_{2}}(z,t)\end{array}\right)+O(\lambda ),\end{eqnarray}$
$\begin{eqnarray}{\phi }_{2}^{\left(1\right)}(z,t,\lambda )=\left(\begin{array}{c}\widetilde{{n}_{1}}(z,t)\\ \widetilde{{n}_{2}}(z,t)\end{array}\right)+O(\lambda ),\end{eqnarray}$
$\begin{eqnarray}{\phi }_{2}^{\left(2\right)}(z,t,\lambda )=\displaystyle \frac{1}{\lambda }\left(\begin{array}{c}{n}_{1}(z,t)\\ {n}_{2}(z,t)\end{array}\right)+O(1).\end{eqnarray}$
Substituting equations ( $\begin{eqnarray}\widetilde{{m}_{1}}(z,t)={\int }_{-\infty }^{z}q(x,t)\widetilde{{m}_{2}}(x,t){\rm{d}}x,\end{eqnarray}$
$\begin{eqnarray}\widetilde{{m}_{2}}(z,t)=1+{\int }_{-\infty }^{z}(\bar{q}(-x,-t)-2\delta )\widetilde{{m}_{1}}(x,t){\rm{d}}x.\end{eqnarray}$
Comparing equations ( $\begin{eqnarray}\left(\begin{array}{c}\widetilde{{m}_{1}}(z,t)\\ \widetilde{{m}_{2}}(z,t)\end{array}\right)=\displaystyle \frac{2{\rm{i}}}{\delta }\left(\begin{array}{c}{m}_{1}(z,t)\\ {m}_{2}(z,t)\end{array}\right).\end{eqnarray}$
Therefore, based on the symmetric condition equation ( $\begin{eqnarray}\left(\begin{array}{c}{n}_{1}(z,t)\\ {n}_{2}(z,t)\end{array}\right)=\left(\begin{array}{c}-\overline{{m}_{2}}(-z,-t)\\ -\overline{{m}_{1}}(-z,-t)\end{array}\right),\end{eqnarray}$
$\begin{eqnarray}\left(\begin{array}{c}\widetilde{{n}_{1}}(z,t)\\ \widetilde{{n}_{2}}(z,t)\end{array}\right)=\left(\begin{array}{c}\overline{\widetilde{{m}_{2}}}(-z,-t)\\ \overline{\widetilde{{m}_{1}}}(-z,-t)\end{array}\right).\end{eqnarray}$
3. The spectral functions
Due to the fact that ${\{{\varphi }_{j}(z,t,\lambda )\}}_{1}^{2}$ are the two matrix solutions to the linear spectral problem (2.7a )–(2.7b ), which are linearly dependent, there exists a scattering matrix G(λ) that is independent of z and t such that2.16 ) into equation (3.1 ) yields2.18 ), we have
$\begin{eqnarray}{\varphi }_{1}(z,t,\lambda )={\varphi }_{2}(z,t,\lambda )G(\lambda ),\quad \lambda \in {\mathbb{R}}\setminus \{0\},\end{eqnarray}$
substituting equation ( $\begin{eqnarray}\begin{array}{rcl}{\phi }_{1}(z,t,\lambda ) & = & {\phi }_{2}(z,t,\lambda )\\ & & \times \,{{\rm{e}}}^{-[{\rm{i}}\lambda z+(4{\rm{i}}\beta {\lambda }^{3}+2{\rm{i}}\alpha {\lambda }^{2})t]\hat{J}}G(\lambda ),\quad \lambda \in {\mathbb{R}}\setminus \{0\},\end{array}\end{eqnarray}$
from the symmetry relation equation ( $\begin{eqnarray}E\overline{{\varphi }_{1}(-z,-t,-\overline{\lambda })}{E}^{-1}={\varphi }_{2}(z,t,\lambda ),\quad \lambda \in {\mathbb{R}}\setminus \{0\}.\end{eqnarray}$
Then, we can define the scattering matrix G(λ) by spectral functions v(λ) and uj(λ), j = 1, 2 as follows $\begin{eqnarray}G(\lambda )=\left(\begin{array}{cc}{u}_{1}(\lambda ) & -\overline{v(-\overline{\lambda })}\\ v(\lambda ) & {u}_{2}(\lambda )\end{array}\right),\quad \lambda \in {\mathbb{R}}\setminus \{0\},\end{eqnarray}$
where u1(λ) and u2(λ) are well-defined in $\lambda \in \overline{{{\mathbb{C}}}_{+}}\setminus \{0\}$ and $\lambda \in \overline{{{\mathbb{C}}}_{-}}\setminus \{0\}$, respectively. Moreover, u1(λ) and u2(λ) have the following symmetry $\begin{eqnarray}\overline{{u}_{1}(-\overline{\lambda })}={u}_{1}(\lambda ),\quad \overline{{u}_{2}(-\overline{\lambda })}={u}_{2}(\lambda ).\end{eqnarray}$
The so-called scattering matrix $G(\lambda )$ given by equation (
Introduce the notations:
$\begin{eqnarray*}\begin{array}{rcl}{\theta }_{11}(z,\lambda ) & = & {\left({\phi }_{1}\right)}_{11}(z,0,\lambda ),\\ {\theta }_{12}(z,\lambda ) & = & {\left({\phi }_{1}\right)}_{12}(z,0,\lambda ),\end{array}\end{eqnarray*}$
$\begin{eqnarray*}\begin{array}{rcl}{\theta }_{21}(z,\lambda ) & = & {\left({\phi }_{1}\right)}_{21}(z,0,\lambda ),\\ {\theta }_{22}(z,\lambda ) & = & {\left({\phi }_{1}\right)}_{22}(z,0,\lambda ).\end{array}\end{eqnarray*}$
From equation ( $\begin{eqnarray}{\theta }_{11}(z,\lambda )=1+{\int }_{-\infty }^{z}{q}_{0}(x){\theta }_{21}(x,\lambda ){\rm{d}}x,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\theta }_{21}(z,\lambda ) & = & \displaystyle \frac{\delta }{2{\rm{i}}\lambda }+{\displaystyle \int }_{-\infty }^{z}{{\rm{e}}}^{2{\rm{i}}\lambda (z-x)}{\theta }_{11}(x,\lambda )({\bar{q}}_{0}(-x)-\delta ){\rm{d}}x\\ & & +\,{\displaystyle \int }_{-\infty }^{z}\displaystyle \frac{\delta }{2{\rm{i}}\lambda }(1-{{\rm{e}}}^{2{\rm{i}}\lambda (z-x)}){q}_{0}(x){\theta }_{21}(x,\lambda ){\rm{d}}x,\end{array}\end{eqnarray}$
$\begin{eqnarray}{\theta }_{12}(z,\lambda )={\int }_{-\infty }^{z}{{\rm{e}}}^{-2{\rm{i}}\lambda (z-x)}{q}_{0}(x){\theta }_{22}(x,\lambda ){\rm{d}}x,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\theta }_{22}(z,\lambda ) & = & 1+{\displaystyle \int }_{-\infty }^{z}({\bar{q}}_{0}(-x)-\delta ){\theta }_{12}(x,\lambda ){\rm{d}}x\\ & & +\,{\displaystyle \int }_{-\infty }^{x}\displaystyle \frac{\delta }{2{\rm{i}}\lambda }({{\rm{e}}}^{-2{\rm{i}}\lambda (z-x)}-1){q}_{0}(x){\theta }_{22}(x,\lambda ){\rm{d}}x.\end{array}\end{eqnarray}$
By solving the systems of these Volterra integral equations, we obtain of the spectral functions $v(\lambda )$ and ${u}_{j}(\lambda )$ $\begin{eqnarray}{u}_{1}(\lambda )=\mathop{\mathrm{lim}}\limits_{z\to \infty }\left({\theta }_{11}(z,\lambda )-\displaystyle \frac{\delta }{2{\rm{i}}\lambda }{\theta }_{21}(z,\lambda )\right),\end{eqnarray}$
$\begin{eqnarray}v(\lambda )=\mathop{\mathrm{lim}}\limits_{z\to \infty }{\theta }_{21}(z,\lambda ){{\rm{e}}}^{-2{\rm{i}}\lambda z},\end{eqnarray}$
$\begin{eqnarray}{u}_{2}(\lambda )=\mathop{\mathrm{lim}}\limits_{z\to \infty }{\theta }_{22}(z,\lambda ).\end{eqnarray}$
These equations indicate that $G(\lambda )$ is uniquely determined by the initial value ${q}_{0}(\lambda )$.The spectral functions $v(\lambda )$ and ${u}_{j}(\lambda ),j=1,2$ can be represented by the following determinant relations:
$\begin{eqnarray}\begin{array}{l}{u}_{1}(\lambda )=\det ({\phi }_{1}^{\left(1\right)}(0,0,\lambda ),{\phi }_{2}^{\left(2\right)}(0,0,\lambda )),\\ \lambda \in \overline{{{\mathbb{C}}}_{+}}\setminus \{0\},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}v(\lambda )=\det ({\phi }_{2}^{\left(1\right)}(0,0,\lambda ),{\phi }_{1}^{\left(2\right)}(0,0,\lambda )),\\ \lambda \in \overline{{{\mathbb{C}}}_{-}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{u}_{2}(\lambda )=\det ({\phi }_{2}^{\left(1\right)}(0,0,\lambda ),{\phi }_{1}^{\left(1\right)}(0,0,\lambda )),\\ \lambda \in {\mathbb{R}}.\end{array}\end{eqnarray}$
The spectral functions $v(\lambda )$ and ${u}_{j}(\lambda ),j=1,2$ have the following properties
(i) ${u}_{1}(\lambda )$ is analytical for $\lambda \in {{\mathbb{C}}}_{+}$ and continue to $\lambda \in \overline{{{\mathbb{C}}}_{+}}\setminus \{0\}$.
(ii) ${u}_{2}(\lambda )$ is analytical for $\lambda \in {{\mathbb{C}}}_{-}$ and continue to $\lambda \in \overline{{{\mathbb{C}}}_{-}}$.
(iii) ${u}_{j}(k)=1\,+\,O(\tfrac{1}{\lambda }),j\,=\,1,2,$ and $v(\lambda )=O(\tfrac{1}{\lambda }),\lambda \to \infty $.
(iv) $\overline{{u}_{1}(-\overline{\lambda })}={u}_{1}(\lambda ),\lambda \in \overline{{{\mathbb{C}}}_{+}}\setminus \{0\},$ and $\overline{{u}_{2}(-\overline{\lambda })}\,={u}_{2}(\lambda ),\lambda \in \overline{{{\mathbb{C}}}_{-}}$.
(v) ${u}_{1}(\lambda ){u}_{2}(\lambda )+v(\lambda )\overline{v(-\overline{\lambda })}=1,\,\lambda \in {\mathbb{R}}\setminus \{0\}$.
(vi) ${u}_{1}(\lambda )=\tfrac{{\delta }^{2}{u}_{2}(0)}{4{\lambda }^{2}}+O(\tfrac{1}{\lambda }),$ and $v(\lambda )=-\tfrac{\delta {u}_{2}(0)}{2{\rm{i}}\lambda }\,+O(1)$, if $\lambda \to 0$.
In fact, the properties (i)-(v) of proposition
$\begin{eqnarray}{u}_{1}(\lambda )=\displaystyle \frac{1}{{\lambda }^{2}}(| {m}_{2}(0,0){| }^{2}-| {m}_{1}(0,0){| }^{2})+O\left(\displaystyle \frac{1}{\lambda }\right),\end{eqnarray}$
$\begin{eqnarray}v(\lambda )=-\displaystyle \frac{2{\rm{i}}}{\lambda \delta }(| {m}_{2}(0,0){| }^{2}-| {m}_{1}(0,0){| }^{2})+O(1),\end{eqnarray}$
$\begin{eqnarray}{u}_{2}(\lambda )=\displaystyle \frac{4}{{\delta }^{2}}(| {m}_{2}(0,0){| }^{2}-| {m}_{1}(0,0){| }^{2})+O(\lambda ),\end{eqnarray}$
from equation ( $\begin{eqnarray}| {m}_{2}(0,0){| }^{2}-| {m}_{1}(0,0){| }^{2}=\displaystyle \frac{{\delta }^{2}{u}_{2}(0)}{4},\end{eqnarray}$
therefore, substituting equation (Take into account that equations (
$\begin{eqnarray}{m}_{2}(z,t)\overline{{m}_{2}}(-z,-t)+{m}_{1}(z,t)\overline{{m}_{1}}(-z,-t)={\rm{const}}.\end{eqnarray}$
If ${q}_{0}(z)$ is the following pure-step initial value:
$\begin{eqnarray}{q}_{0}(z)=\left\{\begin{array}{l}0,\,z\lt 0,\\ \delta ,\,z\gt 0,\end{array}\right.\end{eqnarray}$
then, the so-called scattering matrix $G(\lambda )$ can be given by: $\begin{eqnarray}G(\lambda )={H}_{+}^{-1}(\lambda ){H}_{-}(\lambda )=\left(\begin{array}{cc}1+\tfrac{{\delta }^{2}}{4{\lambda }^{2}} & -\tfrac{\delta }{2{\rm{i}}\lambda }\\ \tfrac{\delta }{2{\rm{i}}\lambda } & 1\end{array}\right).\end{eqnarray}$
4. The basic Riemann–Hilbert problem
To obtain the RH problem of the reverse space-time nonlocal Hirota equation (1.7 ), we need use the Jost function solutions φj(z, t, λ) to define a 2 × 2 matrix-valued function K(z, t, λ) as follows:
$\begin{eqnarray}K(z,t,\lambda )=\left\{\begin{array}{l}\left(\tfrac{{\phi }_{1}^{\left(1\right)}(z,t,\lambda )}{{u}_{1}(\lambda )},{\phi }_{2}^{\left(2\right)}(z,t,\lambda )\right),\lambda \in {{\mathbb{C}}}_{+},\\ \left({\phi }_{2}^{\left(1\right)}(z,t,\lambda ),\tfrac{{\phi }_{1}^{\left(2\right)}(z,t,\lambda )}{{u}_{2}(\lambda )}\right),\lambda \in {{\mathbb{C}}}_{-},\end{array}\right.\end{eqnarray}$
which is a piece-wise meromorphic function relative to ${\mathbb{R}}$ and $\begin{eqnarray}\det K(z,t,\lambda )=1,\quad K(z,t,\lambda )={\mathbb{I}}+O\left(\displaystyle \frac{1}{\lambda }\right),\quad \lambda \to \infty ,\end{eqnarray}$
where ${\mathbb{I}}$ is a 2 × 2 unit matrix.Therefore, the scattering condition equation (3.2 ) means that the boundary values3.3 , we know that the reflection coefficients ξ1(λ) and ξ2(λ) have the symmetry property
$\begin{eqnarray*}\begin{array}{rcl}{K}_{+}(z,t,\lambda ) & = & \mathop{\mathrm{lim}}\limits_{\lambda ^{\prime} \to \lambda ,\,\lambda ^{\prime} \in {{\mathbb{C}}}_{+}}K(z,t,\lambda ^{\prime} ),\quad \lambda \in {\mathbb{R}},\\ {K}_{-}(z,t,\lambda ) & = & \mathop{\mathrm{lim}}\limits_{\lambda ^{\prime} \to \lambda ,\,\lambda ^{\prime} \in {{\mathbb{C}}}_{-}}K(z,t,\lambda ^{\prime} ),\quad \lambda \in {\mathbb{R}},\end{array}\end{eqnarray*}$
enjoys the jump condition as follows $\begin{eqnarray}{K}_{+}(z,t,\lambda )={K}_{-}(z,t,\lambda )S(z,t,\lambda ),\,\lambda \in {\mathbb{R}},\end{eqnarray}$
where S(z, t, λ) is the jump matrix defined by $\begin{eqnarray}S(z,t,\lambda )=\left(\begin{array}{cc}1+{\xi }_{1}(\lambda ){\xi }_{2}(\lambda ) & {\xi }_{2}(\lambda ){{\rm{e}}}^{-2{\rm{i}}\omega (\lambda )}\\ {\xi }_{1}(\lambda ){{\rm{e}}}^{2{\rm{i}}\omega (\lambda )} & 1\end{array}\right),\end{eqnarray}$
and ξj(λ), j = 1, 2 are the reflection coefficients given by $\begin{eqnarray}{\xi }_{1}(\lambda )=\displaystyle \frac{v(\lambda )}{{u}_{1}(\lambda )},\quad {\xi }_{2}(\lambda )=\displaystyle \frac{\overline{v(-\overline{\lambda })}}{{u}_{2}(\lambda )}.\end{eqnarray}$
Moreover, from proposition $\begin{eqnarray}{\xi }_{1}(-\overline{\lambda }){\xi }_{2}(-\overline{\lambda })=\overline{{\xi }_{1}(\lambda )}\overline{{\xi }_{2}(\lambda )},\,\lambda \in {\mathbb{R}}\setminus \{0\},\end{eqnarray}$
and the determinant property $\begin{eqnarray}1+{\xi }_{1}(\lambda ){\xi }_{2}(\lambda )=\displaystyle \frac{1}{{u}_{1}(\lambda ){u}_{2}(\lambda )},\,\lambda \in {\mathbb{R}}\setminus \{0\}.\end{eqnarray}$
The equation (
$\begin{eqnarray}\begin{array}{l}{u}_{1}(\lambda ){\phi }_{2}^{\left(1\right)}(z,t,\lambda )+v(\lambda ){{\rm{e}}}^{2{\rm{i}}\omega (\lambda )}\\ \times \,{\phi }_{2}^{\left(2\right)}(z,t,\lambda )={\phi }_{1}^{\left(1\right)}(z,t,\lambda ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}-\overline{v(-\overline{\lambda }}){{\rm{e}}}^{-2{\rm{i}}\omega (\lambda )}{\phi }_{2}^{\left(1\right)}(z,t,\lambda )+{u}_{2}(\lambda )\\ \times \,{\phi }_{2}^{\left(2\right)}(z,t,\lambda )={\phi }_{1}^{\left(2\right)}(z,t,\lambda ).\end{array}\end{eqnarray}$
These equations and the jump relation equation ( $\begin{eqnarray}\begin{array}{l}\left(\displaystyle \frac{{\phi }_{1}^{\left(1\right)}(z,t,\lambda )}{{u}_{1}(\lambda )},{\phi }_{2}^{\left(2\right)}(z,t,\lambda )\right)\\ \quad =\,\left({\phi }_{2}^{\left(1\right)}(z,t,\lambda ),\displaystyle \frac{{\phi }_{1}^{\left(2\right)}(z,t,\lambda )}{{u}_{2}(\lambda )}\right)S(z,t,\lambda ),\end{array}\end{eqnarray}$
from which, the jump matrix $S(z,t,\lambda )$ defined by equation (Looking back at equations (3.9a )–(3.9c ), it is not difficult to find that the behavior of K(z, t, λ) as λ → 0 mainly depends on whether u2(0) ≠ 0 or u2(0) = 0. Therefore, we have the following assumptions:
Assume that
Case I ${u}_{2}(0)\ne 0$: ${u}_{1}(\lambda )$ has one simple zero located on the imaginary axis in $\overline{{{\mathbb{C}}}_{+}}$, such as $\lambda ={\rm{i}}{\lambda }_{1},{\lambda }_{1}\in {\mathbb{R}}\setminus \{0\};$ ${u}_{2}(\lambda )$ possesses no zeros in $\overline{{{\mathbb{C}}}_{-}}$.
Case II ${u}_{2}(0)=0$: ${u}_{1}(\lambda )$ has one simple zero located on the imaginary axis in $\overline{{{\mathbb{C}}}_{+}}$, such as $\lambda ={\rm{i}}{\lambda }_{1},{\lambda }_{1}\in {\mathbb{R}}\setminus \{0\};$ ${u}_{2}(\lambda )$ possesses one simple zero in $\overline{{{\mathbb{C}}}_{-}}$ at $\lambda =0$ (such that ${\dot{u}}_{2}(0)\ne 0$). Furthermore, we assume that ${\mathrm{lim}}_{\lambda \to 0}\lambda {u}_{1}(\lambda )\ne 0$.
With the symmetric relation equation (
Considering the singularities of ${\{{\phi }_{j}(z,t,\lambda )\}}_{1}^{2}$ and the spectral function u1(λ) at λ = 0 (see proposition 2.1), we can discuss the behavior of K(z, t, λ) at λ = 0 as follows:
In Case I
$\begin{eqnarray}\begin{array}{rcl}{K}_{+}(z,t,\lambda ) & = & \left(\begin{array}{cc}-\displaystyle \frac{4{m}_{1}(z,t)}{{\delta }^{2}{u}_{2}(0)} & -\overline{{m}_{2}}(-z,-t)\\ -\displaystyle \frac{4{m}_{2}(z,t)}{{\delta }^{2}{u}_{2}(0)} & -\overline{{m}_{1}}(-z,-t)\end{array}\right)\\ & & \times ({\mathbb{I}}+O(\lambda ))\left(\begin{array}{cc}\lambda & 0\\ 0 & \displaystyle \frac{1}{\lambda }\end{array}\right),\lambda \to {\rm{i}}0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{K}_{-}(z,t,\lambda ) & = & \displaystyle \frac{2{\rm{i}}}{\delta }\left(\begin{array}{cc}-\overline{{m}_{2}}(-z,-t) & \displaystyle \frac{1}{{u}_{2}(0)}{m}_{1}(z,t)\\ -\overline{{m}_{1}}(-z,-t) & \displaystyle \frac{1}{{u}_{2}(0)}{m}_{2}(z,t)\end{array}\right)\\ & & +\,O(\lambda ),\lambda \to -{\rm{i}}0.\end{array}\end{eqnarray}$
In Case II
$\begin{eqnarray}\begin{array}{rcl}{K}_{+}(z,t,\lambda ) & = & \left(\begin{array}{cc}\displaystyle \frac{{m}_{1}(z,t)}{\mu } & -\overline{{m}_{2}}(-z,-t)\\ \displaystyle \frac{{m}_{2}(z,t)}{\mu } & -\overline{{m}_{1}}(-z,-t)\end{array}\right)\\ & & \times ({\mathbb{I}}+O(\lambda ))\left(\begin{array}{cc}1 & 0\\ 0 & \displaystyle \frac{1}{\lambda }\end{array}\right),\lambda \to {\rm{i}}0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{K}_{-}(z,t,\lambda ) & = & \displaystyle \frac{2{\rm{i}}}{\delta }\left(\begin{array}{cc}-\overline{{m}_{2}}(-z,-t) & \displaystyle \frac{{m}_{1}(z,t)}{{\dot{u}}_{2}(0)}\\ -\overline{{m}_{1}}(-z,-t) & \displaystyle \frac{{m}_{2}(z,t)}{{\dot{u}}_{2}(0)}\end{array}\right)\\ & & \times \,({\mathbb{I}}+O(\lambda ))\left(\begin{array}{cc}1 & 0\\ 0 & \displaystyle \frac{1}{\lambda }\end{array}\right),\lambda \to -{\rm{i}}0.\end{array}\end{eqnarray}$
In addition, if u1(iλ1) = 0 and ${\dot{u}}_{1}({\rm{i}}{\lambda }_{1})\ne 0$ with ${\lambda }_{1}\in {\mathbb{R}}\setminus \{0\}$, then the piece-wise meromorphic function K(z, t, λ) satisfies the following residue condition $\begin{eqnarray}{\mathrm{Res}}_{\lambda ={\rm{i}}{\lambda }_{1}}{K}^{\left(1\right)}(z,t,\lambda )=\displaystyle \frac{{g}_{1}}{{\dot{u}}_{1}({\rm{i}}{\lambda }_{1})}{{\rm{e}}}^{-2{\rm{i}}\omega ({\rm{i}}{\lambda }_{1})}{K}^{\left(2\right)}(z,t,{\rm{i}}{\lambda }_{1}),\end{eqnarray}$
where $\begin{eqnarray*}{\phi }_{1}^{\left(1\right)}(0,0,{\rm{i}}{\lambda }_{1})={g}_{1}{\phi }_{2}^{\left(2\right)}(0,0,{\rm{i}}{\lambda }_{1})\,{and}\,| {g}_{1}| =1.\end{eqnarray*}$
The residue condition equation (
$\begin{eqnarray}\begin{array}{rcl}{\mathrm{Res}}_{\lambda ={\rm{i}}{\lambda }_{1}}\displaystyle \frac{{\phi }_{1}^{\left(1\right)}(z,t,\lambda )}{{u}_{1}(\lambda )} & = & \mathop{\mathrm{lim}}\limits_{\lambda \to {\rm{i}}{\lambda }_{1}}(\lambda -{\rm{i}}{\lambda }_{1})\displaystyle \frac{{\phi }_{1}^{\left(1\right)}(z,t,\lambda )}{{u}_{1}(\lambda )}\\ & = & \displaystyle \frac{{\phi }_{1}^{\left(1\right)}(z,t,\lambda )}{{\dot{u}}_{1}(\lambda )}.\end{array}\end{eqnarray}$
By using ${u}_{1}({\rm{i}}{\lambda }_{1})=0$, we have $\begin{eqnarray}{\phi }_{1}^{\left(1\right)}(z,t,\lambda )=v(\lambda ){{\rm{e}}}^{-2{\rm{i}}\omega ({\rm{i}}{\lambda }_{1})}{K}^{\left(2\right)}(z,t,\lambda ).\end{eqnarray}$
From equations ( $\begin{eqnarray}\begin{array}{l}{\mathrm{Res}}_{\lambda ={\rm{i}}{\lambda }_{1}}{K}^{\left(1\right)}(z,t,\lambda )\\ \quad =\,\displaystyle \frac{v({\rm{i}}{\lambda }_{1})}{{\dot{u}}_{1}({\rm{i}}{\lambda }_{1})}{{\rm{e}}}^{-2{\rm{i}}\omega ({\rm{i}}{\lambda }_{1})}{K}^{\left(2\right)}(z,t,{\rm{i}}{\lambda }_{1}).\end{array}\end{eqnarray}$
these equations will lead to the residue condition equation (Now, let us summarize the results of the above discussion, then we get the following representation theorem.
Assume that ${q}_{0}(z),z\in {\mathbb{R}}$ satisfies the following integral inequality
$\begin{eqnarray*}{\int }_{-\infty }^{0}| {q}_{0}(z)| {\rm{d}}z+{\int }_{0}^{\infty }| {q}_{0}(z)-\delta | {\rm{d}}z\lt \infty .\end{eqnarray*}$
Let $G(\lambda )$ be defined by equation (${u}_{1}(\lambda )$ has a single, simple, pure imaginary zero $\lambda ={\rm{i}}{\lambda }_{1},{\lambda }_{1}\in {\mathbb{R}}\setminus \{0\}$ in $\overline{{{\mathbb{C}}}_{+}}$ (see Case I).
${u}_{2}(\lambda )$ has no zeros in $\overline{{{\mathbb{C}}}_{-}}$ and, if ${u}_{2}(0)=0$, then $\lambda =0$ is a simple zero of ${u}_{2}(\lambda )$ (see Case II).
Consider the following matrix RH problem for the reverse space-time nonlocal Hirota equation (
(i) $K(z,t,\lambda )$ is a piece-wise meromorphic function in ${\mathbb{C}}\setminus {\mathbb{R}}$.
(ii) ${K}_{\pm }(z,t,\lambda )=K(z,t,\lambda \pm {\rm{i}}0),\lambda \in {\mathbb{R}}\setminus \{0\}$ satisfies the jump condition as follows
$\begin{eqnarray*}{K}_{+}(z,t,\lambda )={K}_{-}(z,t,\lambda )S(z,t,\lambda ),\,\lambda \in {\mathbb{R}}\setminus \{0\},\end{eqnarray*}$
where the jump matrix $S(z,t,\lambda )$ is defined in terms of the reflection coefficients ${\xi }_{j}(\lambda ),j\,=\,1,2$ by equation ((iii) $K(z,t,\lambda )={\mathbb{I}}+O(\tfrac{1}{\lambda }),\,\lambda \to \infty $.
(iv) $K(z,t,\lambda )$ satisfies the residue relation equation (
$\begin{eqnarray}q(z,t)=2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }\lambda {K}_{12}(z,t,\lambda ),\end{eqnarray}$
$\begin{eqnarray}\overline{q}(-z,-t)=2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }\lambda {K}_{21}(z,t,\lambda ).\end{eqnarray}$
Therefore, $q(z,t)$ and $\overline{q}(-z,-t)$ solve the reverse space-time nonlocal Hirota equation ( $\begin{eqnarray*}\begin{array}{rcl}{q}_{0}(z) & \to & 0,\,z\to -\infty ,\\ {q}_{0}(z) & \to & \delta ,\,z\to \infty ,\end{array}\end{eqnarray*}$
where ${q}_{0}(z)=q(z,0)$ and $\delta \gt 0$ is a constant.Following [37], we can prove that the solution $K(z,t,\lambda )$ of the RH problem (i)-(iv) defined by theorem 4.5 is unique. Now, we prove equations (
$\begin{eqnarray}\begin{array}{rcl}{\phi }_{\pm }(z,t,\lambda ) & = & {\phi }_{\pm }^{\left(0\right)}+\displaystyle \frac{{\phi }_{\pm }^{\left(1\right)}}{\lambda }+\displaystyle \frac{{\phi }_{\pm }^{\left(2\right)}}{{\lambda }^{2}}\\ & & +\,O(\displaystyle \frac{1}{{\lambda }^{3}}),\lambda \to \infty ,\end{array}\end{eqnarray}$
where ${\phi }_{\pm }^{\left(m\right)}$, $m=0,1,2$ are independent of λ. By using equation ( $\begin{eqnarray}{H}_{\pm }^{-1}(\lambda )={\mathbb{I}}-\displaystyle \frac{1}{2{\rm{i}}\lambda }{Q}_{\pm },\end{eqnarray}$
substituting equations ( $\begin{eqnarray}O({\lambda }^{1}):-{\rm{i}}[J,{\phi }_{\pm }^{\left(0\right)}]=0,\end{eqnarray}$
$\begin{eqnarray}O({\lambda }^{0}):{\phi }_{\pm ,z}^{\left(0\right)}+{\rm{i}}[J,{\phi }_{\pm }^{\left(1\right)}+\displaystyle \frac{{\rm{i}}}{2}{Q}_{\pm }{\phi }_{\pm }^{\left(0\right)}]=(Q-{Q}_{\pm }){\phi }_{\pm }^{\left(0\right)}.\end{eqnarray}$
By using equations (The solution $K(z,t,\lambda )$ of the RH problem (i)-(iv) given by theorem 4.5 has a symmetric relationship as follows:
$\begin{eqnarray}K(z,t,\lambda )=\left\{\begin{array}{l}E\overline{{K}_{+}(-z,-t,-\overline{\lambda })}{E}^{-1}\left(\begin{array}{cc}\tfrac{1}{{u}_{1}(\lambda )} & 0\\ 0 & {u}_{1}(\lambda )\end{array}\right),\,\lambda \in \overline{{{\mathbb{C}}}_{+}}\setminus \{0\},\\ E\overline{{K}_{-}(-z,-t,-\overline{\lambda })}{E}^{-1}\left(\begin{array}{cc}{u}_{2}(\lambda ) & 0\\ 0 & \tfrac{1}{{u}_{2}(\lambda )}\end{array}\right),\,\lambda \in \overline{{{\mathbb{C}}}_{-}}\setminus \{0\}.\end{array}\right.\end{eqnarray}$
From the symmetry of the jump matrix $S(z,t,\lambda )$ defined by equation (
$\begin{eqnarray}\begin{array}{rcl}E\overline{S(-z,-t,-\overline{\lambda })}{E}^{-1} & = & \left(\begin{array}{cc}{u}_{2}(\lambda ) & 0\\ 0 & \tfrac{1}{{u}_{2}(\lambda )}\end{array}\right)S(z,t,\lambda )\\ & & \times \,\left(\begin{array}{cc}\tfrac{1}{{u}_{1}(\lambda )} & 0\\ 0 & {u}_{1}(\lambda )\end{array}\right),\,\lambda \in {\mathbb{R}}\setminus \{0\},\end{array}\end{eqnarray}$
which will lead to the symmetry relation equation (Let ${g}_{1}=1$, we assume that the spectral functions $v(\lambda )$ and ${u}_{j}(\lambda ),j\,=\,1,2$ related to some initial value ${q}_{0}(z)$ meet the following conditions:
$v(\lambda )=0$ for all $\lambda \in {\mathbb{R}}$.
${u}_{2}(\lambda )$ possesses one simple zero in $\overline{{{\mathbb{C}}}_{-}}$ with $\lambda =0$ .
${u}_{1}(\lambda )$ possesses one simple zero in $\overline{{{\mathbb{C}}}_{+}}$ with $\lambda ={\rm{i}}{\lambda }_{1}$, where ${\lambda }_{1}\gt 0$.
Then, ${\lambda }_{1}$ has a unique and definite value ${\lambda }_{1}=\tfrac{\delta }{2}$, and the RH problem (i)-(iv) given by theorem 4.5 has a unique solution, as well as the related exact solution $q(z,t)$ is defined by $\begin{eqnarray}q(z,t)=\displaystyle \frac{\delta }{1-{{\rm{e}}}^{-\delta z-({\rm{i}}\alpha {\delta }^{2}-\beta {\delta }^{3})t}}.\end{eqnarray}$
In fact, it is not difficult to find that the exact solution (4.23 ) is smooth except only one point (0, 0) on (z,t)-plane. Furthermore, it is also easy to see that the exact solution given by equation (4.23 ) satisfies the step-like initial data (1.8a )–(1.8b ).
5. Discussions and conclusions
In this paper, we use the RH method to investigate the Cauchy problem (1.7 )–(1.9b ) of the reverse space-time nonlocal Hirota equation with step-like initial data. Firstly, we present the asymptotic form of Lax pairs (2.12a )–(2.12b ) and their Jost solutions φj(z, t, λ) for the reverse space-time nonlocal Hirota equation. Secondly, we discuss the properties of the scattering matrix G(λ) and the spectral function (the entries of the scattering matrix G(λ)). Finally, we construct the relevant matrix RH problem (see theorem 4.5) and provide a solution to the Cauchy problem (1.7 )–(1.9b ). The theoretical method in this paper can be applied to study the Cauchy problem of other integrable equations with step-like initial data. In addition, the long-time asymptotic analysis of solutions to the Cauchy problem with step-like initial data for nonlocal integrable NLS equation has recently been reported [51]. Therefore, our future research is to discuss these issues.